Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in JAVA):

Solution:
import java.util.*;

public class Main {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int N = Integer.parseInt(sc.next());
        
        double[] p = new double[N];
        
        for(int i = 0; i < N; i++) {
            p[i] = Double.parseDouble(sc.next());
        }
        
        double[][] dp = new double[N+1][N+1]; //コインiまでにj回表
        
        dp[1][0] = 1 - p[0];
        dp[1][1] = p[0];
        
        for(int i = 2; i <= N; i++) {
            for(int j = 0; j <= i; j++) {
                if(j == 0) {
                    dp[i][j] = dp[i-1][j] * (1 - p[i-1]);
                }else if(j < i) {
                    dp[i][j] = dp[i-1][j] * (1 - p[i-1]) + dp[i-1][j-1] * p[i-1];
                }else {
                    dp[i][j] = dp[i-1][j-1] * p[i-1];
                }
            }
        }
        
        double ans = 0;
        for(int j = (N+1) / 2; j <= N; j++) {
            ans += dp[N][j];
        }

        System.out.println(ans);
    }
}